Protein folding reaction

Chemistry Level 2

The folding of a protein can be described by the following reaction equation A B \text{A} \leftrightarrow \text{B} where A is the unfolded chain and B is the folded structure. Under standard conditions, this reaction is accompanied by a change in the free enthalpy of Δ G = 23 kJ / mol \Delta G = -23 \,\text{kJ}/\text{mol} , so that the folded structure is energetically more favorable. How large is the concentration ratio [ B ] [ A] \frac{[\text{B}]}{[\text{A]}} of substances A and B in the reaction equilibrium?

Note: The temperature is T = 2 5 C T = 25^\circ\,\text{C} and the general gas constant is R 8.314 J / ( K mol ) R \approx 8.314 \,\text{J}/(\text{K}\,\text{mol}) .

[ B ] [ A] 1 0 2 \frac{[\text{B}]}{[\text{A]}} \approx 10^2 [ B ] [ A] 1 0 8 \frac{[\text{B}]}{[\text{A]}} \approx 10^8 [ B ] [ A] 1 0 4 \frac{[\text{B}]}{[\text{A]}} \approx 10^4 [ B ] [ A] 1 0 6 \frac{[\text{B}]}{[\text{A]}} \approx 10^6

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Markus Michelmann
Mar 20, 2018

The folding of protein is a first-order reaction because only one molecule is involved at a time. The reaction rates then became d [ A ] d t = k A B [ A ] + k B A [ B ] d [ B ] d t = k B A [ B ] + k A B [ A ] \begin{aligned} \frac{d [A]}{d t} &= - k_{A \to B} [A] + k_{B \to A} [B] \\ \frac{d [B]}{d t} &= - k_{B \to A} [B] + k_{A \to B} [A] \end{aligned} where the coefficients, k A B k_{A \to B} and k B A k_{B \to A} , describe the foward and backward reaction, respectively. In equilibrium, the net reaction rates are zero, so that d [ A ] d t = d [ B ] d t = 0 [ B ] [ A ] = k A B k B A \frac{d [A]}{d t} = \frac{d [B]}{d t} = 0 \quad \Rightarrow \quad \frac{[B]}{[A]} = \frac{k_{A \to B}}{k_{B \to A}} The coefficients follow Arhenius law k A B = C exp ( Δ G A R T ) k B A = C exp ( Δ G A Δ G R T ) \begin{aligned} k_{A \to B} &= C \exp\left( - \frac{\Delta G_A }{R T}\right)\\ k_{B \to A} &= C \exp\left( - \frac{\Delta G_A - \Delta G}{R T}\right) \end{aligned} where Δ G A \Delta G_A is the activation free enthalpy and Δ G < 0 \Delta G < 0 is the reaction free enthalpy.

Finally, the concentration ratio results to [ B ] [ A ] = k A B k B A = exp ( Δ G R T ) = exp ( 23000 8.314 298 ) 1.08 1 0 4 \frac{[B]}{[A]} = \frac{k_{A \to B}}{k_{B \to A}} = \exp\left( - \frac{ \Delta G}{R T}\right) = \exp\left( \frac{23000 }{8.314 \cdot 298}\right) \approx 1.08 \cdot 10^4

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...